Math Problem Statement

A tennis ball is thrown from a cliff at an angle 31 ∘ above the horizontal. The tennis ball hits the ground 3.7 s later at a horizontal distance of 153 ft from where it was thrown. What is the maximum height above the ground that the ball reaches?

Solution

We can approach this problem using the kinematic equations for projectile motion.

Given:

  • Time of flight: t=3.7t = 3.7 s
  • Horizontal distance (range): x=153x = 153 ft
  • Angle of projection: θ=31\theta = 31^\circ
  • Gravity: g=32.2ft/s2g = 32.2 \, \text{ft/s}^2 (acceleration due to gravity)

We aim to find the maximum height of the ball above the ground.

Step 1: Horizontal Motion

The horizontal velocity vxv_x is constant, and the horizontal distance traveled is given by:

x=vxtx = v_x \cdot t

The horizontal velocity vxv_x is related to the initial velocity v0v_0 and the angle of projection θ\theta as:

vx=v0cos(θ)v_x = v_0 \cdot \cos(\theta)

Substituting into the equation for the range:

153=v0cos(31)3.7153 = v_0 \cdot \cos(31^\circ) \cdot 3.7

We can solve for v0v_0.

Step 2: Vertical Motion

The vertical component of the initial velocity is:

vy=v0sin(θ)v_y = v_0 \cdot \sin(\theta)

The maximum height is reached when the vertical velocity becomes zero. Using the kinematic equation:

vy2=v0y22ghmaxv_y^2 = v_{0y}^2 - 2g h_{\text{max}}

At the maximum height, vy=0v_y = 0, so:

0=v02sin2(θ)2ghmax0 = v_0^2 \cdot \sin^2(\theta) - 2g h_{\text{max}}

Solving for hmaxh_{\text{max}}:

hmax=v02sin2(31)2gh_{\text{max}} = \frac{v_0^2 \cdot \sin^2(31^\circ)}{2g}

We will first solve for v0v_0 from Step 1, then calculate hmaxh_{\text{max}}. Let's compute.The initial velocity of the tennis ball is approximately 48.24ft/s48.24 \, \text{ft/s}.

The maximum height the ball reaches above the point it was thrown from is approximately 9.59ft9.59 \, \text{ft}.

If you'd like further details or have any questions, feel free to ask!

Here are some related questions to explore:

  1. How do the horizontal and vertical components of the velocity affect the motion?
  2. What would happen if the angle of projection was increased?
  3. How does air resistance affect projectile motion, and why is it often ignored in such calculations?
  4. Can you calculate the time taken to reach the maximum height?
  5. What is the velocity of the tennis ball when it hits the ground?

Tip: When dealing with projectile motion, it's useful to break down the motion into horizontal and vertical components to simplify calculations.

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Math Problem Analysis

Mathematical Concepts

Projectile Motion
Kinematics
Trigonometry

Formulas

Horizontal distance: x = v_x * t
Horizontal velocity: v_x = v_0 * cos(θ)
Vertical velocity: v_y = v_0 * sin(θ)
Kinematic equation for maximum height: v_y^2 = v_{0y}^2 - 2g * h_max

Theorems

Projectile Motion Theorems

Suitable Grade Level

Grades 11-12